When is a polynomial \(p(x,y)\) expressible in the form \[ p(x,y)=f(x+iy)+g(x-iy), \] where \(f\) and \(g\) are holomorphic?
Solution
Such a representation exists exactly when \(p\) is harmonic: \[ \boxed{\Delta p=p_{xx}+p_{yy}=0.} \] Equivalently, after rewriting \(p\) uniquely as a polynomial in \(z=x+iy\) and \(\overline z=x-iy\), no monomial \(z^j\overline z^{\,k}\) with \(j,k>0\) may occur.
1. If\[ p(x,y)=f(z)+g(\overline z), \]with \(f\) and \(g\) holomorphic, then \(p\) is harmonic.
Using Wirtinger derivatives, \[ \partial_{\overline z}f(z)=0, \qquad \partial_z g(\overline z)=0. \] Hence \[ \partial_z\partial_{\overline z}p=0. \] Since \[ \Delta=\partial_x^2+\partial_y^2 =4\partial_z\partial_{\overline z}, \] we obtain \(\Delta p=0\).
2. Rewrite \(p\) uniquely as\[ p(x,y)=P(z,\overline z) =\sum_{j,k\ge0}c_{jk}z^j\overline z^{\,k}. \]
The linear change of variables \[ z=x+iy, \qquad \overline z=x-iy \] is invertible, with \[ x=\frac{z+\overline z}{2}, \qquad y=\frac{z-\overline z}{2i}. \] Therefore substitution gives an isomorphism \[ \mathbb C[x,y]\cong\mathbb C[z,\overline z] \] of polynomial rings, and the displayed expansion is unique.
3. If \(p\) is harmonic, then \(c_{jk}=0\) whenever \(j,k>0\).
By harmonicity and <1>1’s identity for the Laplacian, \[ 0=\partial_z\partial_{\overline z}P =\sum_{j,k\ge1}jk\,c_{jk} z^{j-1}\overline z^{\,k-1}. \] Distinct monomials in the polynomial ring \(\mathbb C[z,\overline z]\) are linearly independent, so \[ jk\,c_{jk}=0 \] for every \(j,k\ge1\). Since \(jk\neq0\), all such \(c_{jk}\) vanish.
4. Every harmonic polynomial therefore has the required form.
By 3, \[ P(z,\overline z) =c_{00} +\sum_{j\ge1}c_{j0}z^j +\sum_{k\ge1}c_{0k}\overline z^{\,k}. \] Define the one-variable polynomials \[ f(w)=c_{00}+\sum_{j\ge1}c_{j0}w^j, \qquad g(w)=\sum_{k\ge1}c_{0k}w^k. \] They are entire, and \[ p(x,y)=f(x+iy)+g(x-iy). \] Together with 1, this proves the equivalence.